The function provided in the image is:

$f(x) = \frac{x-7}{(x-5)(x+4)}$

To determine where $f(x)$ is continuous, we need to identify the values of $x$ that cause the denominator to be zero since these will be the points of discontinuity.

The denominator is zero when:

$(x-5)(x+4) = 0$

Solving this equation, we find:

$x-5 = 0 \quad \text{or} \quad x+4 = 0$ $x = 5 \quad \text{or} \quad x = -4$

Therefore, the function $f(x)$ is discontinuous at $x = 5$ and $x = -4$. Elsewhere, the function is continuous since the numerator and denominator are polynomials, and polynomials are continuous everywhere.

So, the function $f(x)$ is continuous for all $x$ except at $x = 5$ and $x = -4$.

Would you like to explore how the limits behave at these points or have other questions on continuity or limits?

Here are five related questions you might consider:

1. What is the behavior of $f(x)$ as $x$ approaches 5?
2. How does $f(x)$ behave as $x$ approaches -4?
3. Can $f(x)$ be redefined at $x = 5$ or $x = -4$ to make it continuous at these points?
4. What is the limit of $f(x)$ as $x$ approaches infinity?
5. How would the graph of $f(x)$ look around the points of discontinuity?

Tip: When checking for continuity of a rational function, it’s helpful to factorize the denominator and look for common factors in the numerator, which can simplify the expression and potentially remove some points of discontinuity.